Particles in a gas lose and gain energy at random due to collisions with each other. On average, over a large number of particles, the proportion of particles which have at least a certain amount of energy ε is constant. This is known as the Boltzmann factor. It is a value between 0 and 1. The Boltzmann factor is given by the formula:

${\displaystyle {\frac {n}{n_{0}}}=e^{\frac {-\epsilon }{kT}}}$,

where n is the number of particles with kinetic energy above an energy level ε, n0 is the total number of particles in the gas, T is the temperature of the gas (in kelvin) and k is the Boltzmann constant (1.38 x 10-23 JK-1).

This energy could be any sort of energy that a particle can have - it could be gravitational potential energy, or kinetic energy, for example.

Derivation

In the atmosphere, particles are pulled downwards by gravity. They gain and lose gravitational potential energy (mgh) due to collisions with each other. First, let's consider a small chunk of the atmosphere. It has horizontal cross-sectional area A, height dh, molecular density (the number of molecules per. unit volume) n and all the molecules have mass m. Let the number of particles in the chunk be N.

${\displaystyle n={\frac {N}{V}}={\frac {N}{A\;dh}}}$

Because:

${\displaystyle V=A\;dh}$ (which makes sense, if you think about it)

By definition:

${\displaystyle N=nV=nA\;dh}$

The total mass Σ m is the mass of one molecule (m) multiplied by the number of molecules (N):

${\displaystyle \Sigma m=mN=mnA\;dh}$

Then work out the weight of the chunk:

${\displaystyle W=g\Sigma m=nmgA\;dh}$

The downwards pressure P is force per. unit area, so:

${\displaystyle P={\frac {W}{A}}={\frac {nmgA\;dh}{A}}=nmg\;dh}$

We know that, as we go up in the atmosphere, the pressure decreases. So, across our little chunk there is a difference in pressure dP given by:

${\displaystyle dP=-nmg\;dh}$ (1) In other words, the pressure is decreasing (-) and it is the result of the weight of this little chunk of atmosphere.

We also know that:

${\displaystyle PV=NkT}$

So:

${\displaystyle P={\frac {NkT}{V}}}$

But:

${\displaystyle n={\frac {N}{V}}}$

So, by substitution:

${\displaystyle P=nkT}$

So, for our little chunk:

${\displaystyle dP=kT\;dn}$ (2)

If we equate (1) and (2):

${\displaystyle dP=-nmg\;dh=kT\;dn}$

Rearrange to get:

${\displaystyle {\frac {dn}{dh}}={\frac {-nmg}{kT}}}$

${\displaystyle {\frac {dh}{dn}}={\frac {-kT}{nmg}}}$

Integrate between the limits n0 and n:

${\displaystyle h={\frac {-kT}{mg}}\int _{n_{0}}^{n}{\frac {1}{n}}\;dn={\frac {-kT}{mg}}\left[\ln {n}\right]_{n_{0}}^{n}={\frac {-kT}{mg}}\left(\ln {n}-\ln {n_{0}}\right)={\frac {-kT}{mg}}\ln {\frac {n}{n_{0}}}}$

${\displaystyle ln{\frac {n}{n_{0}}}={\frac {-mgh}{kT}}}$

${\displaystyle {\frac {n}{n_{0}}}=e^{\frac {-mgh}{kT}}}$

Since we are dealing with gravitational potential energy, ε = mgh, so:

${\displaystyle {\frac {n}{n_{0}}}=e^{\frac {-\epsilon }{kT}}}$

A Graph of this Function

This topic comes up in Q10 494 June 2010. The Values used for various things in that question are

${\displaystyle k=1.4\times 10^{-23}JK^{-1},~g=9.8,~m=4.9\times 10^{-26}Kg,~T=290K}$

Shows how Energies are achieved with Height

Questions

1u = 1.66 x 10-27 kg

g = 9.81 ms-2

1. A nitrogen molecule has a molecular mass of 28u. If the Earth's atmosphere is 100% nitrous, with a temperature of 18°C, what proportion of nitrogen molecules reach a height of 2km?

2. What proportion of the molecules in a box of hydrogen (molecular mass 2u) at 0°C have a velocity greater than 5ms-1?

3. What is the temperature of the hydrogen if half of the hydrogen is moving at at least 10ms-1?

4. Some ionised hydrogen (charge -1.6 x 10-19 C)is placed in a uniform electric field. The potential difference between the two plates is 20V, and they are 1m apart. What proportion of the molecules are at least 0.5m from the positive plate (ignoring gravity) at 350°K?

Worked Solutions